Saved Bookmarks
| 1. |
(a) The current density in a cylindrical wire of radius R=2.0mm is uniform across a cross section of the wire and is J=2.0 xx 10^(5) A//m^(2). What is the current through the outer portion of the wire between radial distances R/2 and R? (b) Suppose, instead, the current density through a cross section varies with radial distance r as J=ar^(2), in which a=3.0 xx 10^(11)A//m^(4) and r is in meters. What now is the current through the same outer portion of the wire? |
|
Answer» Solution :(a) Because the current density is UNIFORM across the cross section, the current density J, the current I, and the cross-sectional AREA A are related by (J=i/A). Calculation: We want only the current through a reduced cross-sectional area A. of the wire (rather than the entire area), where `A.=pi R^(2)-pi (R/2)^(2)=pi ((3R^(2))/(4))` `=(3pi)/(4) (0.0020 m)^(2)=9.424 xx 10^(-6) m^(2)`. So, we REWRITE Eq. 26-5 as i=JA. and then substitute the data to find `i=(2.0 xx 10^(5) A//m^(2)) (9.424 xx 10^(-6) m^(2))` =1.9A (b) Because the current density is not uniform across a cross section of the wire, we must resort to `(i=int vecJ.d vecA)` and integrate the current density over the portion of the wire from r=R/2 to r=R. Calculations: The current density vector `vecJ` (along the wire.s length) and the differential area vector dA (perendicular to a cross section of the wire) have the same direction. Thus, `vecJ=d vecA=JdA cos 0=jdA`. we need to replace the differential area dA with something we can actually integrate between the limits r=R/2 and r=R. The simplest replacement (because J is given as a function of f) is the area of `2pir` of a thin RING of circumference `2pir` and width dr. We can then integrate with r as the variable of integration. Eq 26-4 then gives as `i=int vecJ.d vecA=int JdA` `=int_(R//2)^(R) ar^(2) 2pir dr=2pi a int_(R/2)^(R) r^(2) dr` `=2pia [r^(4)/]_(R//2)^(R) =(pi a)/(2) [R^(4)-(R^(4)/(16))]=(15)/(32) piaR^(4)` `=(15)/(32) pi (3.0 xx 10^(11) A//m^(4)) (0.0020 m)^(4)=7.1A` |
|